1. **State the problem:** Simplify the logical expression $$[p \leftrightarrow (q \lor r)] \leftrightarrow [\neg (p \to (r \to q)) \lor \neg (\neg q \land \neg r) \to p].$$ 2. **Recall logical equivalences:** - Implication: $$a \to b \equiv \neg a \lor b$$ - Biconditional: $$a \leftrightarrow b \equiv (a \land b) \lor (\neg a \land \neg b)$$ - De Morgan's laws: $$\neg (a \land b) \equiv \neg a \lor \neg b$$ and $$\neg (a \lor b) \equiv \neg a \land \neg b$$ 3. **Simplify inner implications:** - $$p \to (r \to q) \equiv \neg p \lor (\neg r \lor q) \equiv \neg p \lor \neg r \lor q$$ 4. **Negate the implication:** - $$\neg (p \to (r \to q)) \equiv \neg (\neg p \lor \neg r \lor q) \equiv p \land r \land \neg q$$ 5. **Simplify $$\neg (\neg q \land \neg r)$$:** - By De Morgan's law: $$\neg (\neg q \land \neg r) \equiv q \lor r$$ 6. **Rewrite the right side:** - $$\neg (p \to (r \to q)) \lor \neg (\neg q \land \neg r) \to p \equiv (p \land r \land \neg q) \lor (q \lor r) \to p$$ - Using implication: $$a \to p \equiv \neg a \lor p$$, so - $$[(p \land r \land \neg q) \lor (q \lor r)] \to p \equiv \neg [(p \land r \land \neg q) \lor (q \lor r)] \lor p$$ 7. **Negate the disjunction:** - $$\neg [(p \land r \land \neg q) \lor (q \lor r)] \equiv \neg (p \land r \land \neg q) \land \neg (q \lor r)$$ - $$\neg (p \land r \land \neg q) \equiv \neg p \lor \neg r \lor q$$ - $$\neg (q \lor r) \equiv \neg q \land \neg r$$ - So the negation is $$(\neg p \lor \neg r \lor q) \land (\neg q \land \neg r)$$ 8. **Simplify the conjunction:** - $$(\neg p \lor \neg r \lor q) \land \neg q \land \neg r$$ - Since $$\neg r$$ and $$\neg q$$ are true, the disjunction simplifies: - $$\neg p \lor \neg r \lor q \equiv \neg p \lor \text{false} \lor \text{false} \equiv \neg p$$ - So the whole expression is $$\neg p \land \neg q \land \neg r$$ 9. **Add the $$\lor p$$ from step 6:** - $$\neg [(p \land r \land \neg q) \lor (q \lor r)] \lor p \equiv (\neg p \land \neg q \land \neg r) \lor p$$ 10. **Rewrite the entire right side:** - $$[(p \land r \land \neg q) \lor (q \lor r)] \to p \equiv (\neg p \land \neg q \land \neg r) \lor p$$ 11. **Rewrite the left side:** - $$p \leftrightarrow (q \lor r) \equiv (p \land (q \lor r)) \lor (\neg p \land \neg (q \lor r))$$ - $$\neg (q \lor r) \equiv \neg q \land \neg r$$ - So left side is $$(p \land (q \lor r)) \lor (\neg p \land \neg q \land \neg r)$$ 12. **Final expression:** - $$[(p \land (q \lor r)) \lor (\neg p \land \neg q \land \neg r)] \leftrightarrow [(\neg p \land \neg q \land \neg r) \lor p]$$ 13. **Check equivalence:** - Both sides contain $$(\neg p \land \neg q \land \neg r)$$ - The right side also has $$p$$, the left side has $$p \land (q \lor r)$$ - When $$p$$ is true and $$(q \lor r)$$ is false, left side is false, right side is true, so expression is false. 14. **Conclusion:** The expression is not a tautology and cannot be simplified further to a simpler form without context. **Final simplified form:** $$[(p \land (q \lor r)) \lor (\neg p \land \neg q \land \neg r)] \leftrightarrow [(\neg p \land \neg q \land \neg r) \lor p].$$